Question

求解一道关于极限的题目

f(x)=x^2+x,a(n+1)=f(an),a1=0.5
求lim[1/(1+a1)+1/(1+a2)+......+1/(1+an)] (n趋向于无穷）
{an}为一数列

Answer 1

a(n+1)=f(an)=an^2+an
1/a(n+1)=1/(an^2+an)=1/[an(1+an)]=1/an - 1/(1+an)
1/(1+an)=1/an-1/a(n+1)
1/(1+a1)+1/(1+a2)+......+1/(1+an)
=(1/a1-1/a2)+(1/a2-1/a3)+...[1/an-1/a(n+1)]
=1/a1-1/a(n+1)
=2-1/a(n+1)

a(n+1)=an^2+an
a(n+1)-an=an^2>0 (a1>0)
a(n+1)>an>a(n-1)>...>a1=1/2
a2-a1=a1^2
a3-a2=a2^2
......
an-a(n-1)=a(n-1)^2（上述式子进行叠加）
an-a1=a1^2+a2^2+a3^2+...+a(n-1)^2
an=a1+a1^2+a2^2+a3^2+...+a(n-1)^2
>1/2+(1/2)^2+(1/2)^2+(1/2)^2+...+(1/2)^2 （共有n-1个(1/2)^2 ）
=1/2+(n-1)/4
=(n+1)/4
a(n+1)>[(n+1)+1]/4=(n+2)/4
1/a(n+1)<4/(n+2)
当n趋向于无穷,4/(n+2)趋向于0,即1/a(n+1)趋向于0
lim[1/(1+a1)+1/(1+a2)+......+1/(1+an)]
=lim[2-1/a(n+1)]
=2